Integration by parts problem involving vector functions

In summary, the conversation discusses the proof of reciprocity in MRI and the use of integration by parts to simplify the expression for \Phi_{M}. The conversation also mentions the use of a vector identity and explains the concept of ignoring surface term for finite current sources.
  • #1
vabamyyr
66
0
Hi,

I am trying to chew through the proof of reciprocity in MRI. At some point I come across to the following expression:

[itex]\Phi_{M}[/itex]=[itex]\oint\vec{dl}[/itex][itex]\cdot[/itex][itex]\left[\frac{\mu_{0}}{4\pi}\int{d^{3}r'}\frac{\vec{\nabla'}\times\vec{M}(\vec{r'})}{\left|\vec{r}-\vec{r'}\right|}\right][/itex]

Now it says that by using integration by parts (where surface term can be ignored for finite current sources), and using vector identity, [itex]\vec{A}\cdot\left(\vec{B}\times\vec{C}\right)=-\left(\vec{A}\times\vec{C}\right)\cdot\vec{B}[/itex], we get

[itex]\Phi_{M}=\frac{\mu_{0}}{4\pi}\int{d^{3}r'\vec{M}(\vec{r'})\cdot\left[\vec{\nabla'}\times\left(\oint\frac{\vec{dl}}{\left|\vec{r}-\vec{r'}\right|}\right)\right]}[/itex]

Can someone explain to me how to use integration by parts in this case and what does ignoring surface term mean in this context?
 
Physics news on Phys.org
  • #2
vabamyyr said:
Hi,

I am trying to chew through the proof of reciprocity in MRI. At some point I come across to the following expression:

[itex]\Phi_{M}[/itex]=[itex]\oint\vec{dl}[/itex][itex]\cdot[/itex][itex]\left[\frac{\mu_{0}}{4\pi}\int{d^{3}r'}\frac{\vec{\nabla'}\times\vec{M}(\vec{r'})}{\left|\vec{r}-\vec{r'}\right|}\right][/itex]

Now it says that by using integration by parts (where surface term can be ignored for finite current sources), and using vector identity, [itex]\vec{A}\cdot\left(\vec{B}\times\vec{C}\right)=-\left(\vec{A}\times\vec{C}\right)\cdot\vec{B}[/itex], we get

[itex]\Phi_{M}=\frac{\mu_{0}}{4\pi}\int{d^{3}r'\vec{M}(\vec{r'})\cdot\left[\vec{\nabla'}\times\left(\oint\frac{\vec{dl}}{\left|\vec{r}-\vec{r'}\right|}\right)\right]}[/itex]

Can someone explain to me how to use integration by parts in this case and what does ignoring surface term mean in this context?
This link is helpful.
 

Related to Integration by parts problem involving vector functions

1. What is integration by parts?

Integration by parts is a mathematical technique used to evaluate integrals of products of two functions. It is based on the product rule of differentiation and allows us to transform a difficult integral into a simpler one.

2. How is integration by parts used with vector functions?

Integration by parts can be used with vector functions by considering each component of the vector as a separate function. The integration by parts formula is then applied to each component separately, and the results are combined to obtain the final solution.

3. What are the steps involved in solving an integration by parts problem with vector functions?

The steps for solving an integration by parts problem with vector functions are as follows:
1. Identify the functions u and v in the integral.
2. Use the product rule to find the derivatives of u and v.
3. Substitute the derivatives into the integration by parts formula.
4. Evaluate the resulting integrals.
5. Combine the results to obtain the final solution.

4. Are there any special cases to consider when using integration by parts with vector functions?

Yes, there are two special cases to consider:
1. If one of the functions in the integral is a constant, then the integration by parts formula simplifies.
2. If one of the functions is a trigonometric function, then a substitution may be necessary to simplify the integral.

5. Can integration by parts be used with higher dimensions?

Yes, integration by parts can be extended to higher dimensions by using the concept of vector calculus. In this case, the integration by parts formula becomes a surface integral, and the product rule is replaced by the divergence theorem or Green's theorem.

Similar threads

  • Calculus
Replies
2
Views
1K
Replies
2
Views
2K
Replies
2
Views
1K
Replies
5
Views
1K
  • Electromagnetism
Replies
14
Views
1K
Replies
1
Views
1K
Replies
1
Views
513
Replies
1
Views
491
  • Calculus and Beyond Homework Help
Replies
9
Views
840
Back
Top